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Proof step 3 from proof divides-sum (previous step)

∀ i. (∀ j. j ∈ ints0 → i | j) → i | Σ ints0, ∀ j. j = i0 ∨ j ∈ ints0 → i | j ⊦ i | i0 + Σ ints0

User applied rule all left

And obtained the following new goals

Proof step 4:
∀ i. (∀ j. j ∈ ints0 → i | j) → i | Σ ints0, i0 = i0 ∨ i0 ∈ ints0 → i | i0, ∀ j. j ∈ ints0 → i | j ⊦ i | i0 + Σ ints0

Rule argument

pos: left 2 terms: i0 abort: F discard: F computed: T

Additional information

Used terms: i0